Show that if $Q$ is orthogonal transformation matrix, then $Q^t(Q-1)=(1-Q)^t$. Deduce that if $Q$ is also proper, then $\det(1-Q)=0$. Hence show that transformation has nonzero vector that has the same components in both coordinate system.

I tried to solve this problem.I think I got the first part right,

$$Q^t(Q -1)= Q^t Q- Q^t=1- Q^t=(1- Q)^t$$

The second part,

$$-Q ^t(1-Q)=(1-Q)^t$$



since the orthogonal matrix is proper which means $\det(Q)=1$ and for any matrix, its determinant equals the determinant of its transpose.


So, it's always true for $\det(1-Q)=0$

But that's not what the question asks. I haven't done linear algebra for a while and I am not sure from the concepts I used, so I would be glad if you clarify any mistake I made.

  • $\begingroup$ Unrelated: which context did this come from? I've almost never seen $\lambda$ used to denote a matrix instead of an eigenvalue before. $\endgroup$ – YiFan Feb 15 at 23:44
  • $\begingroup$ Me neither. In linear algebra, I always see λ denotes an eigenvalue, but this is how my professor wrote it. I will edit it to avoid any misunderstanding. $\endgroup$ – K.ali Feb 15 at 23:56
  • $\begingroup$ not mentioned in the question. $\endgroup$ – K.ali Feb 16 at 0:15

Since $Q$ satisfies $Q^t(Q-1)=(1-Q)^t$, then $\det(Q)\det(Q-1)=\det(1-Q)=(-1)^n\det(Q-1)$, as you observed. Suppose $\det(Q-1)$ is nonzero, then we can divide both sides by it to get $\det Q=(-1)^n$. If $n$ is odd, we get a contradiction with the fact that $Q$ is proper.

The statement is false when $n$ is even. Consider $\mathbb R^2$ and $Q=-1=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$. We have $\det Q=1$, but $\det(1-Q)=4\neq0$.

For the last part, we need to show the existence of $x$ in the vector space so that $Qx=x$, i.e. one of the eigenvalues of $Q$ is $1$. But we've just done that!


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